\(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\cdot\dfrac{49}{100}=\dfrac{98}{100}>\dfrac{1}{4}\)

A = \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)

A = \(2.\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\right)\)

A = \(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

A = \(2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

A = \(1-\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\)

~ Chúc bạn học giỏi ! ~

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)

\(\Rightarrow2A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{100}\)

\(\Rightarrow A=1-\dfrac{1}{50}\)

\(\Rightarrow A=\dfrac{49}{50}\)

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)

\(\dfrac{1}{2}A=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\right)\)

\(\dfrac{1}{2}A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)

\(\dfrac{1}{2}A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{100}\)

\(\dfrac{1}{2}A=\dfrac{49}{100}\)

\(A=\dfrac{49}{50}\)

Ta có công thức tổng quát sau:

\(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)

\(=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Ta có: \(\left(1-\frac13\right)\left(1-\frac16\right)\cdot\ldots\cdot\left(1-\frac{1}{4950}\right)\)

\(=\left(1-\frac26\right)\left(1-\frac{2}{12}\right)\cdot\ldots\cdot\left(1-\frac{2}{9900}\right)\)

\(=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot...\cdot\left(1-\frac{2}{99\cdot100}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(99+2\right)\left(99-1\right)}{99\left(99+1\right)}\)

\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{101\cdot98}{99\cdot100}\)

\(=\frac{4\cdot5\cdot\ldots\cdot101}{3\cdot4\cdot\ldots\cdot100}\cdot\frac{1\cdot2\cdot\ldots\cdot98}{2\cdot3\cdot\ldots\cdot99}=\frac{101}{3}\cdot\frac{1}{99}=\frac{101}{297}\)

x3x

D=1/3+1/6+1/10+1/15+......+1/4950

=2x（1/6+1/12+1/20+1/30+……+1/9900）

=2x（1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+……+1/99-1/100）

=2x（1/2-1/100）

=1-1/50

=49/50

**** nhé

A=1+3+6+10+...+4851+4950
2A=2+6+12+20+...+9702+9900
2A=1.2+2.3+3.4+4.5+...+98.99+99.100
Xét B=1.2+2.3+3.4+4.5+...+98.99+99.100
3B=1.2.3+2.3(4−1)+3.4(5−2)+...+99.100(101−98)
3B=1.2.3+2.3.4−1.2.3+3.4.5−2.3.4+...+99.100.101−98.99.100
3B=99.100.101
B=333300
Thay B vào A ta được:
2A=333300
A=166650 

1/3+1/6+1/10+1/15+......+1/4950

=2x（1/6+1/12+1/20+1/30+……+1/9900）

=2x（1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+……+1/99-1/100）

=2x（1/2-1/100）

=1-1/50

=49/50

**** nhé

Đáp án là 49/50 nha hatsune miku 

Chúc bạn học tốt!